How do you simplify sin(tan^-1(x))?
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Question “How do you simplify sin(tan^-1(x))?”
How do you simplify sin(tan^-1(x))?
Answer
Answer 1
sin(arctan(x)) = |x|/sqrt(x^2+1)
Explanation:
This is what you need to know
sin^2(theta) + cos^2(theta) = 1
We divide both sides by
sin^2(theta)
1 + cot^2(theta) = csc^2(theta)
Or,
1 + 1/tan^2(theta) = 1/sin^2(theta)
Take the least-common multiple that we have
(tan^2(theta) + 1)/tan^2(theta) = 1/sin^2(theta)
We have both the positive and negative sides.
sin^2(theta) = tan^2(theta)/(tan^2(theta) + 1)
We say this.
theta = Arctan(x).
sin^2(arctan(x)) = tan^2(arctan(x))/(tan^2(arctan(x)) + 1)
This is what you need to know
tan(arctan(x)) = x
sin^2(arctan(x)) = x^2/(x^2 + 1)
We take the square root from both sides.
sin(arctan(x)) = +-sqrt(x^2/(x^2+1)) = +-|x|/sqrt(x^2+1)
We can see that the sine is always positive when we check the range of the Arctangent.
sin(arctan(x)) = |x|/sqrt(x^2+1)
Answer 2
sin(tan^-1(x))=x/sqrt(x^2+1)
Explanation:
The principles of “SOH CAH-TOA” can be used.
Let’s start by calling
sin(tan^-1(x))=sin(theta)
theta=tan^-1(x)
More precisely,
tan^-1(x)=theta
tan(theta)=x
Since
tan(theta)=”opposite”/”adjacent”
tan(theta)=x/1
{(“opposite”=x),(“adjacent”=1),(“hypotenuse”=? ):}
Use
The Pythagorean Theory
We can see that the hypotenuse is a right triangle with legs.
x
1
“hypotenuse”=sqrt(x^2+1)
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sin(tan^-1(x))
sin(theta)
(“opposite”=x),(“adjacent”=1),(“hypotenuse”=sqrt(x^2+1)):
Since
sin(theta)=”opposite”/”hypotenuse”
sin(tan^-1(x))=x/sqrt(x^2+1)
Conclusion
Above is the solution for “How do you simplify sin(tan^-1(x))?“. We hope that you find a good answer and gain the knowledge about this topic of trigonometry.